The energy associated with Bohr's orbit in the hydrogen atom is given by the expression, `E_(n) = - (13.6)/(n^(2))eV.` the energy in eV
associated with the orbit having a radius 9 `r_(1)` is (`r_(1)` is the radius of the first orbit )

Options

A. `-13.6`
B. `-6.8`
C. `-15.1`
D. `-1.36`

Answer

C

Explanation

Given,
Energy associated with Bohr.s orbit in the hydrogen atom `E_(n) = - (13.6)/(n^(2)) ` eV
now, radius of Bohr.s orbit, ` r_(n)` = 0.529 `(n^(2))/(Z)` Å
or `" " r_(n) = r_(1) n^(2)` `" " [because Z =1 " for hydrogen "]`
Given, `r_(n) = 9r_(1)`
or `" " r_(1) n^(2) = r_(1) 9`
or `" " n^(2) = 9 `
or `" " ` n = 3
So, the energy associated with orbit of radius `9r_(1)` can be given as :
` E_(9) = - (13.6)/(9) eV = - 1.51` eV
or `" " E_(9) = - 1.51 ` eV